The decay constant of a radio isotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1$ and $

Question

The decay constant of a radio isotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1$ and $t_2$ respectively, the number of nuclei which have decayed during the time $(t_1 - t_2)$ is:

Accepted Answer

1. **Definition of Activity:** The activity ($A$) of a radioactive substance is defined as the rate of disintegration of nuclei. It is related to the number of undecayed nuclei ($N$) present at that instant and the decay constant ($\lambda$) by the relation: $A = \lambda N$ .
2. **Relate N to A:** From the relation, we can express the number of nuclei as $N = \frac{A}{\lambda}$.
3. **Initial and Final States:**
   - At time $t_1$, activity is $A_1$, so the number of nuclei present is $N_1 = \frac{A_1}{\lambda}$.
   - At time $t_2$, activity is $A_2$, so the number of nuclei present is $N_2 = \frac{A_2}{\lambda}$.
4. **Calculate Decayed Nuclei:** The number of nuclei that have decayed during the interval is the difference between the initial number of nuclei and the final number of nuclei.
   $$ 	ext{Number decayed} = N_1 - N_2 $$
   $$ 	ext{Number decayed} = \frac{A_1}{\lambda} - \frac{A_2}{\lambda} = \frac{A_1 - A_2}{\lambda} $$

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